Linear Independence/Dependence of set

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Homework Statement



Suppose that x,y, and z are distinct vectors in a vector space V over a field F, and S = {x,y,z} is linearly independent. If S* = span({x+z, x-y}), prove whether S* is linearly independent or linearly dependent.

Homework Equations





The Attempt at a Solution



S* = a(x+z) + b(x-y) = ax + bx -by + az = (a+b)x - by + az, where a and b are coefficients. We know that this linear relationship only has the trivial solution (because we are told that the set S = {x,y,z} is linearly independent), thus S* must be linearly independent.


This is the solution I have, but apparently it's wrong. The set is linearly dependent. Why is that?

EDIT: I know that I didn't write a linear combination of the span, I just wrote out the span. But the span itself is a linear combination of the vectors and a linear combination of a linear combination is still a linear combination, so just to simplify things I only wrote out the span of the vectors.
 
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Answers and Replies

  • #2
Office_Shredder
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Is S* the whole span of those vectors, or is S* just the two vectors? If it's the span, it's obviously linearly dependent, as e.g. x+z, 2(x+z) are both in S*, so a(x+z) + b[2(x+z)] = 0 has b=-1, a=2 as a solution
 
  • #3
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S* is the span. But the two vectors are (x+z) and (x-y). Not (x+z) and (x+z).
 
  • #4
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Wow, I just realized what you were really trying to say. :rofl:

I see what you are saying now. But what is wrong with my proof?
 
  • #5
Office_Shredder
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Your proof is that the two vectors x+z and x-y are linearly independent, which they are. The set of all vectors of the form a*(x+z) + b*(x-y) isn't... to check this, you'd take a look at a summation

[tex]\sum_{i=1}^n(a_i(x+z) + b_i(x-y)) = 0[/tex] for any scalars ai and bi (n is an arbitrary finite natural number)
 
  • #6
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Thanks. I think I understand now (your first example made it easy)

Let me give my own example to see if I really have it down.

Let's say v = [tex]a_1(x+z) + b_1(x-y))[/tex] and

p = [tex]a_2(x+z) + b_2(x-y)) [/tex]

Then if you choose [tex]a_2 = -a_1[/tex] and [tex]b_2 = -b_1[/tex], then obviously the set consisting of v and p is linearly dependent.
 
  • #7
Dick
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Sure, that's right. But you are still thinking too hard. I like (x+z) and 2(x+z) much better. Because it shows you how obvious it really is. The span contains an infinite numbers of vectors. A linearly independent set over a two dimensional subspace (like the span) contains only two. Even a subset containing three vectors MUST be linearly dependent. An infinite number is way overkill.
 

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